Assuming ideality:
PV=nRT
Pv= (mass/molarmass)RT
Solve for the mass :-)
I first converted the given grams of the reactants into moles, and then divided the moles by the coefficients in front of each of the reactant. The result with the smallest value will be the limiting reactant, and the value of CuO was the smallest, so it's the limiting reactant.
After figuring out which reactant is the limiting one, I took their given grams and converted it into moles, the divided it by the ratio of N2 to CuO (it's in the equation) to obtain the moles of N2, and then multiply it with the molar mass of N2 to get its mass in grams.
Answer:
Yes, it is possible.
Explanation:
A diprotic acid is an acid that can release two protons. That's why it is called diprotic.
Monoprotic → Release one proton, for example Formic acid HCOOH
Triprotic → Releases three protons, for example H₃PO₄
Polyprotic → Release many protons, for example EDTA
it is a weak acid.
In the first equilibrum, it release proton, and the second is released in the second equilibrium. So the first equilibrium will have a Ka1
H₂A + H₂O ⇄ H₃O⁺ + HA⁻ Ka₁
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
The HA⁻ will work as an amphoterous because, it can be a base or an acid, according to this:
HA⁻ + H₂O ⇄ H₃O⁺ + A⁻² Ka₂
HA⁻ + H₂O ⇄ OH⁻ + H₂A Kb₂
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